chapter 12 - j. falabella portfolio€¦ · 12.3 limiting reagent and percent yield > 10...

12.3 Limiting Reagent and Percent Yield >

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Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield



CHEMISTRY & YOU

What determines how much product you can make?

If a carpenter had two tabletops and seven table legs, he would have difficulty building more than one functional four-legged table.



Limiting and Excess Reagents

Limiting and Excess Reagents How is the amount of product in a reaction affected by an insufficient quantity of any of the reactants?




To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings.




To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. •  If you have only 2 taco shells, the quantity of

taco shells will limit the number of tacos you can make.




To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. •  If you have only 2 taco shells, the quantity of

taco shells will limit the number of tacos you can make.

•  Thus, the taco shells are the limiting ingredient.




In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.




A balanced chemical equation is a chemist’s recipe. Chemical Equations

N2(g) + 3H2(g) 2NH3(g)

“Microscopic recipe” 1 molecule N2 + 3 molecules H2 2 molecules NH3

“Macroscopic recipe” 1 mol N2 + 3 mol H2 2 mol NH3




A balanced chemical equation is a chemist’s recipe.

•  What would happen if two molecules (moles) of N2 reacted with three molecules (moles) of H2?

Experimental Conditions Reactants Products

Before reaction

2 molecules N2 3 molecules H2

0 molecules NH3

Chemical Equations N2(g) + 3H2(g) 2NH3(g)

“Microscopic recipe” 1 molecule N2 + 3 molecules H2 2 molecules NH3

“Macroscopic recipe” 1 mol N2 + 3 mol H2 2 mol NH3




•  Before the reaction takes place, N2 and H2 are present in a 2:3 molecule (mole) ratio.

•  As the reaction takes place, one molecule (mole) of N2 reacts with 3 molecules (moles) of H2 to produce two molecules (moles) of NH3.


Before reaction


0 molecules NH3

After reaction

1 molecule N2 0 molecules H2

2 molecules NH3





Before reaction


0 molecules NH3

After reaction


2 molecules NH3

•  All the H2 has now been used up, and the reaction stops.

•  One molecule (mole) of unreacted N2 is left in addition to the two molecules (moles) of NH3 that have been produced by the reaction.





Before reaction


0 molecules NH3

After reaction


2 molecules NH3

•  In this reaction, only the hydrogen is completely used up.

•  H2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction.





Before reaction


0 molecules NH3

After reaction


2 molecules NH3

•  The reactant that is not completely used up in a reaction is called the excess reagent.

•  In this example, nitrogen is the excess reagent because some nitrogen remains unreacted.



CHEMISTRY & YOU

What determines how much product you can make in a chemical reaction?



CHEMISTRY & YOU

What determines how much product you can make in a chemical reaction?

A limited quantity of any of the reactants that are needed to make a product will limit the amount of product that forms.



Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:

2Cu(s) + S(s) → Cu2S(s)

What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

Sample Problem 12.8

Determining the Limiting Reagent in a Reaction



Analyze List the knowns and the unknown. 1 The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant.

KNOWNS mass of copper = 80.0 g Cu

mass of sulfur = 25.0 g S

molar mass of Cu = 63.5 g/mol

molar mass of S = 32.1 g/mol

1 mol S/2 mol Cu

UNKNOWN limiting reagent = ?

Sample Problem 12.8



Start with one of the reactants and convert from mass to moles.

Calculate Solve for the unknown. 2

Sample Problem 12.8

80.0 g Cu × = 1.26 mol Cu 63.5 g Cu 1 mol Cu



Then, convert the mass of the other reactant to moles.


Sample Problem 12.8

25.0 g S × = 0.779 mol S 32.1 g S 1 mol S



Now, convert moles of Cu to moles of S needed to react with 1.25 moles of Cu.


Sample Problem 12.8

1.26 mol Cu × = 0.630 mol S 2 mol Cu 1 mol S

Given quantity

Mole ratio

Needed amount




Sample Problem 12.8

Compare the amount of sulfur needed with the given amount of sulfur.

0.630 mol S (amount needed to react) <0.779 mol S (given amount)

Sulfur is in excess, so copper is the limiting reagent.

It doesn’t matter which reactant you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent.



Since the ratio of the given mol Cu to mol S was less than the ratio (2:1) from the balanced equation, copper should be the limiting reagent.

Evaluate Do the results make sense? 3

Sample Problem 12.8



What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S?

2Cu(s) + S(s) → Cu2S(s)

Sample Problem 12.9

Using Limiting Reagent to Find the Quantity of a Product



Analyze List the knowns and the unknown. 1 The limiting reagent, which was determined in the previous sample problem, is used to calculate the maximum amount of Cu2S formed.

KNOWNS limiting reagent = 1.26 mol Cu (from sample problem 12.8)

1 mol Cu2S = 159.1 g Cu2S (molar mass)

1 mol Cu2S/2 mol Cu (mole ratio from balanced equation)

UNKNOWN Yield = ? g Cu2S

Sample Problem 12.9



Start with the moles of the limiting reagent and convert to moles of the product. Use the mole ratio from the balanced equation.


Sample Problem 12.9

1.26 mol Cu × 2 mol Cu 1 mol Cu2S



Finish the calculation by converting from moles to mass of product.


Sample Problem 12.9

159.1 g Cu2S 1 mol Cu2S

1.26 mol Cu × 2 mol Cu 1 mol Cu2S

×

= 1.00 × 102 g Cu2S



•  Copper is the limiting reagent in this reaction. The maximum number of grams of Cu2S produced should be more than the amount of copper that initially reacted because copper is combining with sulfur.

•  However, the mass of Cu2S produced should be less than the total mass of the reactants (105.0 g) because sulfur was in excess.

Evaluate Do the results make sense? 3

Sample Problem 12.9



Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is

2Fe + O2 + 2H2O → 2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?



Only 2.26 g H2O are needed to react with 7.0 g Fe. Therefore, Fe is the limiting reagent.

7.00 g Fe × × × = 2.26 g H20 1 mol Fe

55.85 g Fe 18.0 g H2O 1 mol H2O 2 mol Fe

2 mol H2O

Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is

2Fe + O2 + 2H2O → 2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?



Percent Yield

Percent Yield What does the percent yield of a reaction measure?



Percent Yield

Percent Yield What does the percent yield of a reaction measure?

•  A batting average is actually a percent yield.



Percent Yield

When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield.



Percent Yield

When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield. •  The theoretical yield is the maximum amount

of product that could be formed from given amounts of reactants.

•  The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.



Percent Yield

The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.

Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%.

percent yield = actual yield

theoretical yield × 100%



The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.

Percent Yield



The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.

Percent Yield

The mass of the

reactant is measured.

The mass of one of the products, the actual yield, is measured. The percent yield is calculated.

The reactant is heated.



Percent Yield

Many factors cause percent yields to be less than 100%. •  Reactions do not always go to completion; when a

reaction is incomplete, less than the calculated amount of product is formed.

•  Impure reactants and competing side reactions may cause unwanted products to form.

•  Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers.

•  If reactants or products have not been carefully measured, a percent yield of 100% is unlikely.



Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is

Sample Problem 12.10

Calculating the Theoretical Yield of a Reaction

CaCO3(s) → CaO(s) + CO2(g) Δ

What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?



Analyze List the knowns and the unknown. 1 Calculate the theoretical yield using the mass of the reactant.

KNOWNS mass of CaCO3 = 24.8 g CaCO3

1 mol CaCO3 = 100.1 g CaCO3 (molar mass)

1 mol CaO = 56.1 g CaO (molar mass)

1 mol CaO/1 mol CaCO3 (mole ratio from balanced equation)

UNKNOWN theoretical yield = ? g CaO




Start with the mass of the reactant and convert to moles of the reactant.



24.8 g CaCO3 × 100.1 g CaCO3 1 mol CaCO3



Next, convert to moles of the product using the mole ratio.



24.8 g CaCO3 × × 100.1 g CaCO3 1 mol CaCO3

1 mol CaCO3 1 mol CaO



Finish by converting from moles to mass of the product.



24.8 g CaCO3 × × × 100.1 g CaCO3 1 mol CaCO3

1 mol CaCO3 1 mol CaO

1 mol CaO 56.1 g CaO

= 13.9 g CaO If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction.



•  The mole ratio of CaO to CaCO3 is 1:1. The ratio of their masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater than 1:2.

•  The result of the calculations shows that the mass of CaO is slightly greater than half the mass of CaCO3.

Evaluate Does the result make sense? 3




What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated?


Calculating the Percent Yield of a Reaction

CaCO3(s) → CaO(s) + CO2(g) Δ

Calculate the theoretical yield first. Then you can calculate the percent yield.



Analyze List the knowns and the unknown. 1 Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem 12.10.

UNKNOWN percent yield = ? %


KNOWNS actual yield = 13.1 g CaO

theoretical yield = 13.9 g CaO (from sample problem 12.10)





Substitute the values for actual yield and theoretical yield into the equation for percent yield.



percent yield = × 100% = 94.2% 13.1 g CaO 13.9 g CaO



•  In this example, the actual yield is slightly less than the theoretical yield.

•  Therefore, the percent yield should be slightly less than 100%.

Evaluate Does the result make sense? 3




Carbon tetrachloride, CCl4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is

CS2 + 3Cl2 → CCl4 + S2Cl2 What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?



CS2 + 3Cl2 → CCl4 + S2Cl2

What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?

3.12 × 105 g CS2 × × × 76.142 g CS2 1 mol CS2 1 mol CCl4

1 mol CS2 153.81 g CCl4

1 mol CCl4

= 6.30 × 105 g CCl4 = 630 kg CCl4

Percent yield = × 100% = 97.9% 617 kg CCl4 630 kg CCl4



Key Concepts and Key Equation

In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

The percent yield is a measure of the efficiency of a reaction performed in the laboratory.





Glossary Terms

•  limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction

•  excess reagent: a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction



•  theoretical yield: the amount of product that could form during a reaction calculated from a balanced chemical equation; it represents the maximum amount of product that could be formed from a given amount of reactant

•  actual yield: the amount of product that forms when a reaction is carried out in the laboratory

•  percent yield: the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage; a measure of the efficiency of a reaction

Glossary Terms



The Mole and Quantifying Matter

BIG IDEA

The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction.



END OF 12.3

chapter 12 - j. falabella portfolio€¦ · 12.3 limiting reagent and percent yield > 10...

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